/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  a : [] --> o 
  b : [] --> o 
  f : [o * o] --> o 

  f(a, f(b, X)) => f(a, f(a, f(a, X))) 
  f(b, f(a, X)) => f(b, f(b, f(b, X))) 

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]).

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] f#(a, f(b, X)) =#> f#(a, f(a, f(a, X)))   
    1] f#(a, f(b, X)) =#> f#(a, f(a, X))   
    2] f#(a, f(b, X)) =#> f#(a, X)   
    3] f#(b, f(a, X)) =#> f#(b, f(b, f(b, X)))   
    4] f#(b, f(a, X)) =#> f#(b, f(b, X))   
    5] f#(b, f(a, X)) =#> f#(b, X)   

  Rules R_0:

    f(a, f(b, X)) => f(a, f(a, f(a, X))) 
    f(b, f(a, X)) => f(b, f(b, f(b, X))) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0, 1, 2 
    * 1 : 0, 1, 2 
    * 2 : 0, 1, 2 
    * 3 : 3, 4, 5 
    * 4 : 3, 4, 5 
    * 5 : 3, 4, 5 

This graph has the following strongly connected components:

  P_1:

    f#(a, f(b, X)) =#> f#(a, f(a, f(a, X)))   
    f#(a, f(b, X)) =#> f#(a, f(a, X))   
    f#(a, f(b, X)) =#> f#(a, X)   

  P_2:

    f#(b, f(a, X)) =#> f#(b, f(b, f(b, X)))   
    f#(b, f(a, X)) =#> f#(b, f(b, X))   
    f#(b, f(a, X)) =#> f#(b, X)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_2, R_0, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  f#(b, f(a, X)) >? f#(b, f(b, f(b, X))) 
  f#(b, f(a, X)) >? f#(b, f(b, X)) 
  f#(b, f(a, X)) >? f#(b, X) 
  f(a, f(b, X)) >= f(a, f(a, f(a, X))) 
  f(b, f(a, X)) >= f(b, f(b, f(b, X))) 

We orient these requirements with a polynomial interpretation in the natural numbers.

We consider usable_rules with respect to the following argument filtering:


This leaves the following ordering requirements:

  f#(b, f(a, X)) >= f#(b, f(b, f(b, X))) 
  f#(b, f(a, X)) > f#(b, f(b, X)) 
  f#(b, f(a, X)) >= f#(b, X) 
  f(b, f(a, X)) >= f(b, f(b, f(b, X))) 

The following interpretation satisfies the requirements:

  a = 3 
  b = 0 
  f = \y0y1.y1 + 3y0 
  f# = \y0y1.2y1 

Using this interpretation, the requirements translate to:

  [[f#(b, f(a, _x0))]] = 18 + 2x0 > 2x0 = [[f#(b, f(b, f(b, _x0)))]] 
  [[f#(b, f(a, _x0))]] = 18 + 2x0 > 2x0 = [[f#(b, f(b, _x0))]] 
  [[f#(b, f(a, _x0))]] = 18 + 2x0 > 2x0 = [[f#(b, _x0)]] 
  [[f(b, f(a, _x0))]] = 9 + x0 >= x0 = [[f(b, f(b, f(b, _x0)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_0, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  f#(a, f(b, X)) >? f#(a, f(a, f(a, X))) 
  f#(a, f(b, X)) >? f#(a, f(a, X)) 
  f#(a, f(b, X)) >? f#(a, X) 
  f(a, f(b, X)) >= f(a, f(a, f(a, X))) 
  f(b, f(a, X)) >= f(b, f(b, f(b, X))) 

We orient these requirements with a polynomial interpretation in the natural numbers.

We consider usable_rules with respect to the following argument filtering:


This leaves the following ordering requirements:

  f#(a, f(b, X)) >= f#(a, f(a, f(a, X))) 
  f#(a, f(b, X)) > f#(a, f(a, X)) 
  f#(a, f(b, X)) >= f#(a, X) 
  f(a, f(b, X)) >= f(a, f(a, f(a, X))) 

The following interpretation satisfies the requirements:

  a = 0 
  b = 3 
  f = \y0y1.y1 + 3y0 
  f# = \y0y1.2y1 

Using this interpretation, the requirements translate to:

  [[f#(a, f(b, _x0))]] = 18 + 2x0 > 2x0 = [[f#(a, f(a, f(a, _x0)))]] 
  [[f#(a, f(b, _x0))]] = 18 + 2x0 > 2x0 = [[f#(a, f(a, _x0))]] 
  [[f#(a, f(b, _x0))]] = 18 + 2x0 > 2x0 = [[f#(a, _x0)]] 
  [[f(a, f(b, _x0))]] = 9 + x0 >= x0 = [[f(a, f(a, f(a, _x0)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.